\(z\) is Odd in Any Primitive Pythagorean Triple \((x, y, z)\)

Theorem

Let \((x, y, z)\) by a primitive pythagorean triple, that is

\[ x^2 + y^2 = z^2\]

and \(\gcd(x, y, z) = 0\).

Then, \(z\) must be odd.

Proof

Let \((x, y, z)\) be a primitive Pythagorean triple, in which \(z\) is even.

That is, let \(z\) be either \(0\) or \(2\) modulo \(4\). Therefore \(z^2 \equiv 0 \pmod 4\).

Hence,

\[ x^2 + y^2 \equiv 0 \pmod 4.\]

Up to reordering, the possible equivalence classes for \(x\) and \(y\) modulo \(4\) are therefore

\[ (0, 0), (1,3), \quad \text{and} \quad (2, 2).\]

However, in the cases of \((0, 0)\) and \((2, 2)\), \(x\), \(y\) and \(z\) would be even, which contradicts the assumption that the triple is primitive, yet it is impossible for \(x^2\) or \(y^2\) to be \(3\) modulo \(4\), as is obvious when computing quadratic residues modulo \(4\) explicitly

\[\begin{align*} 0^2 &\equiv 0 \pmod 4 \\ 1^2 &\equiv 1 \pmod 4 \\ 2^2 &\equiv 0 \pmod 4 \\ 3^2 &\equiv 1 \pmod 4 \\ \end{align*}\]